Is $1^\frac{1}{3}$ ambiguous?

Question

I know that the three 3rd roots of unity are $1$, $(-1)^\frac{2}{3}$, and $-(-1)^\frac{1}{3}$. I also believe that $1^\frac{1}{3}$ is a cube root of unity since $(1^\frac{1}{3})^3=1$, and that $1^\frac{1}{3}= (-1)^\frac{2}{3}$ since, \begin{equation} 1^\frac{1}{3}= (-e^{i\pi})^\frac{1}{3} = (-1)^\frac{1}{3}(e^{i\pi})^\frac{1}{3}=(-1)^\frac{1}{3}(-1)^\frac{1}{3}=(-1)^\frac{2}{3}. \end{equation} However, when solving $x^3=1$, I find $x=\sqrt[3]{1}=1^\frac{1}{3}$ which has those three unique roots as solutions. Does $\sqrt[3]{1}\neq1^\frac{1}{3}$? Or, is $1^\frac{1}{3}$ ambiguous in some way that $(-1)^\frac{2}{3}$ is not?

Answer 1

It's a matter of definitions. For example, in the education that I received, the definitions were the following:

• Surd function $\sqrt[n]x$, $n\in\mathbb{N^+}$, $x\in\mathbb{R}$ if $n$ is odd, $x\ge 0$ if $n$ is even. By definition, $y=\sqrt[n]x$ is the largest real root of $y^n = x$. So $\sqrt 4 = 2$, $\sqrt[3]{-1} = -1$ and $\sqrt{-1}$ is undefined

• Power function with real exponent $x^p$, $p\in\mathbb{R}$, $x\ge 0$ if $p\neq 0$, $x>0$ if $p=0$. By definition, $y=x^p=\lim_{\frac{n}{m}\to p}\sqrt[m]{x^n}$. Here, $1^{1/3}=1$, $(-1)^{1/3}$ is undefined, $0^0$ is undefined too. The reason to limit scope to only non-negative numbers is because you want to preserve the property $a^pb^p=(ab)^p$ and $(a^p)^q = a^{pq}$. Note, that $\sqrt[n]{x}$ isn't generally the same as $x^{1/n}$.

• Complex exponent $e^z=\exp{z}=1+z+z^2/2+...$, $z\in\mathbb C$. Note, that with this set of definition $e^z$ isn't generally the same as “number $e$ to the power of $z$”. Since the latter is only defined for real $z$.

By this set of definitions, $1^{1/3}=1$ and isn't ambiguous. However, your expression $(-e^{i\pi})^{1/3}=(-1)^{1/3}e^{i\pi/3}$ is wrong because power function $x^y$ is not defined for negative argument $x=-1$.

What options do you have if you still want to define something like $(-1)^{2/3}$?

1. As in some software, you can set $x^y=e^{y\ln x}$ where $\ln x$ is a principal value of a logarithm. However, you are losing the power properties I listed above (that's exactly what happened $(-e^{i\pi})^{1/3}\neq(-1)^{1/3}e^{i\pi/3}$ because by this definition generally $x^ay^a\neq(xy)^a$)and also losing continuity: $$ (-1+\epsilon i)^{1/2} \approx i, \qquad (-1-\epsilon i)^{1/2} \approx -i,\qquad\text{for small $\epsilon>0$} $$

2. You can define $y=x^{p/q}$ as a set of roots $y^q=x^p$, you will retain continuity on $x$, but it only really works for rational powers (or you have to deal with infinite sets). And you still have problems with $(x^a)^b$ property: $(-1)^{1/3}$ is a set of 3 elements and $(-1)^{3/9}$ is a set of 9.

In general, I would avoid using power notation if the argument is not non-negative or the exponent is not real. Otherwise, you need to explicitly include your definition of power function if you don't want to be misunderstood.